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# Largest subarray with equal number of 0s and 1s

• Difficulty Level : Hard
• Last Updated : 18 Jun, 2021

Given an array containing only 0s and 1s, find the largest subarray which contains equal no of 0s and 1s. The expected time complexity is O(n).

Examples:

```Input: arr[] = {1, 0, 1, 1, 1, 0, 0}
Output: 1 to 6
(Starting and Ending indexes of output subarray)

Input: arr[] = {1, 1, 1, 1}
Output: No such subarray

Input: arr[] = {0, 0, 1, 1, 0}
Output: 0 to 3 Or 1 to 4```

Method 1: Brute Force.

Approach: The brute force approach in these type of questions is to generate all the possible sub-arrays. Then firstly check whether the sub-array has equal number of 0’s and 1’s or not. To make this process easy take cumulative sum of the sub-arrays taking 0’s as -1 and 1’s as it is. The point where cumulative sum = 0 will signify that the sub-array from starting till that point has equal number of 0’s and 1’s. Now as this is a valid sub-array, compare it’s size with the maximum size of such sub-array found till now.

Algorithm :

1. Use a starting a pointer which signifies the starting point of the sub-array.
2. Take a variable sum=0 which will take the cumulative sum of all the sub-array elements.
3. Initialize it with value 1 if the value at starting point=1 else initialize it with -1.
4. Now start an inner loop and start taking the cumulative sum of elements following the same logic.
5. If the cumulative sum (value of sum)=0 it signifies that the sub-array has equal number of 0’s and 1’s.
6. Now compare its size with the size of the largest sub-array if it is greater store the first index of such sub-array in a variable and update the value of size.
7. Print the sub-array with the starting index and size returned by the above algorithm.

Pseudo Code:

```Run a loop from i=0 to n-1
if(arr[i]==1)
sum=1
else
sum=-1
Run inner loop from j=i+1 to n-1
sum+=arr[j]
if(sum==0)
if(j-i+1>max_size)
start_index=i
max_size=j-i+1
Run a loop from i=start_index till max_size-1
print(arr[i])```

## C++

 `// A simple C++ program to find the largest``// subarray with equal number of 0s and 1s``#include ` `using` `namespace` `std;` `// This function Prints the starting and ending``// indexes of the largest subarray with equal``// number of 0s and 1s. Also returns the size``// of such subarray.` `int` `findSubArray(``int` `arr[], ``int` `n)``{``    ``int` `sum = 0;``    ``int` `maxsize = -1, startindex;` `    ``// Pick a starting point as i``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``sum = (arr[i] == 0) ? -1 : 1;` `        ``// Consider all subarrays starting from i``        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``(arr[j] == 0) ? (sum += -1) : (sum += 1);` `            ``// If this is a 0 sum subarray, then``            ``// compare it with maximum size subarray``            ``// calculated so far``            ``if` `(sum == 0 && maxsize < j - i + 1) {``                ``maxsize = j - i + 1;``                ``startindex = i;``            ``}``        ``}``    ``}``    ``if` `(maxsize == -1)``        ``cout << ``"No such subarray"``;``    ``else``        ``cout << startindex << ``" to "``             ``<< startindex + maxsize - 1;` `    ``return` `maxsize;``}` `/* Driver code*/``int` `main()``{``    ``int` `arr[] = { 1, 0, 0, 1, 0, 1, 1 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSubArray(arr, size);``    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `// A simple program to find the largest subarray``// with equal number of 0s and 1s` `#include ` `// This function Prints the starting and ending``// indexes of the largest subarray with equal``// number of 0s and 1s. Also returns the size``// of such subarray.` `int` `findSubArray(``int` `arr[], ``int` `n)``{``    ``int` `sum = 0;``    ``int` `maxsize = -1, startindex;` `    ``// Pick a starting point as i` `    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``sum = (arr[i] == 0) ? -1 : 1;` `        ``// Consider all subarrays starting from i` `        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``(arr[j] == 0) ? (sum += -1) : (sum += 1);` `            ``// If this is a 0 sum subarray, then``            ``// compare it with maximum size subarray``            ``// calculated so far` `            ``if` `(sum == 0 && maxsize < j - i + 1) {``                ``maxsize = j - i + 1;``                ``startindex = i;``            ``}``        ``}``    ``}``    ``if` `(maxsize == -1)``        ``printf``(``"No such subarray"``);``    ``else``        ``printf``(``"%d to %d"``, startindex, startindex + maxsize - 1);` `    ``return` `maxsize;``}` `/* Driver program to test above functions*/` `int` `main()``{``    ``int` `arr[] = { 1, 0, 0, 1, 0, 1, 1 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSubArray(arr, size);``    ``return` `0;``}`

## Java

 `class` `LargestSubArray {` `    ``// This function Prints the starting and ending``    ``// indexes of the largest subarray with equal``    ``// number of 0s and 1s. Also returns the size``    ``// of such subarray.` `    ``int` `findSubArray(``int` `arr[], ``int` `n)``    ``{``        ``int` `sum = ``0``;``        ``int` `maxsize = -``1``, startindex = ``0``;``        ``int` `endindex = ``0``;` `        ``// Pick a starting point as i` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``            ``sum = (arr[i] == ``0``) ? -``1` `: ``1``;` `            ``// Consider all subarrays starting from i` `            ``for` `(``int` `j = i + ``1``; j < n; j++) {``                ``if` `(arr[j] == ``0``)``                    ``sum += -``1``;``                ``else``                    ``sum += ``1``;` `                ``// If this is a 0 sum subarray, then``                ``// compare it with maximum size subarray``                ``// calculated so far` `                ``if` `(sum == ``0` `&& maxsize < j - i + ``1``) {``                    ``maxsize = j - i + ``1``;``                    ``startindex = i;``                ``}``            ``}``        ``}``        ``endindex = startindex + maxsize - ``1``;``        ``if` `(maxsize == -``1``)``            ``System.out.println(``"No such subarray"``);``        ``else``            ``System.out.println(startindex + ``" to "` `+ endindex);` `        ``return` `maxsize;``    ``}` `    ``/* Driver program to test the above functions */` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``LargestSubArray sub;``        ``sub = ``new` `LargestSubArray();``        ``int` `arr[] = { ``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1` `};``        ``int` `size = arr.length;` `        ``sub.findSubArray(arr, size);``    ``}``}`

## Python3

 `# A simple program to find the largest subarray``# with equal number of 0s and 1s` `# This function Prints the starting and ending``# indexes of the largest subarray with equal``# number of 0s and 1s. Also returns the size``# of such subarray.``def` `findSubArray(arr, n):` `    ``sum` `=` `0``    ``maxsize ``=` `-``1` `    ``# Pick a starting point as i` `    ``for` `i ``in` `range``(``0``, n``-``1``):``    ` `        ``sum` `=` `-``1` `if``(arr[i] ``=``=` `0``) ``else` `1` `        ``# Consider all subarrays starting from i` `        ``for` `j ``in` `range``(i ``+` `1``, n):``        ` `            ``sum` `=` `sum` `+` `(``-``1``) ``if` `(arr[j] ``=``=` `0``) ``else` `sum` `+` `1` `            ``# If this is a 0 sum subarray, then``            ``# compare it with maximum size subarray``            ``# calculated so far` `            ``if` `(``sum` `=``=` `0` `and` `maxsize < j``-``i ``+` `1``):``                ` `                ``maxsize ``=` `j ``-` `i ``+` `1``                ``startindex ``=` `i``            ` `        ` `    ` `    ``if` `(maxsize ``=``=` `-``1``):``        ``print``(``"No such subarray"``);``    ``else``:``        ``print``(startindex, ``"to"``, startindex ``+` `maxsize``-``1``);` `    ``return` `maxsize` `# Driver program to test above functions``arr ``=` `[``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1``]``size ``=` `len``(arr)``findSubArray(arr, size)` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// A simple program to find the largest subarray``// with equal number of 0s and 1s``using` `System;` `class` `GFG {` `    ``// This function Prints the starting and ending``    ``// indexes of the largest subarray with equal``    ``// number of 0s and 1s. Also returns the size``    ``// of such subarray.` `    ``static` `int` `findSubArray(``int``[] arr, ``int` `n)``    ``{``        ``int` `sum = 0;``        ``int` `maxsize = -1, startindex = 0;``        ``int` `endindex = 0;` `        ``// Pick a starting point as i``        ``for` `(``int` `i = 0; i < n - 1; i++) {``            ``sum = (arr[i] == 0) ? -1 : 1;` `            ``// Consider all subarrays starting from i` `            ``for` `(``int` `j = i + 1; j < n; j++) {``                ``if` `(arr[j] == 0)``                    ``sum += -1;``                ``else``                    ``sum += 1;` `                ``// If this is a 0 sum subarray, then``                ``// compare it with maximum size subarray``                ``// calculated so far` `                ``if` `(sum == 0 && maxsize < j - i + 1) {``                    ``maxsize = j - i + 1;``                    ``startindex = i;``                ``}``            ``}``        ``}``        ``endindex = startindex + maxsize - 1;``        ``if` `(maxsize == -1)``            ``Console.WriteLine(``"No such subarray"``);``        ``else``            ``Console.WriteLine(startindex + ``" to "` `+ endindex);` `        ``return` `maxsize;``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{` `        ``int``[] arr = { 1, 0, 0, 1, 0, 1, 1 };``        ``int` `size = arr.Length;``        ``findSubArray(arr, size);``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output:

` 0 to 5`

Complexity Analysis:

• Time Complexity: O(n^2).
As all the possible sub-arrays are generated using a pair of nested loops.
• Auxiliary Space: O(1).
As no extra data structure is used which takes auxiliary space.

Method 2: Hashmap.

Approach: The concept of taking cumulative sum, taking 0’s as -1 will help us in optimizing the approach. While taking the cumulative sum, there are two cases when there can be a sub-array with equal number of 0’s and 1’s.

1. When cumulative sum=0, which signifies that sub-array from index (0) till present index has equal number of 0’s and 1’s.
2. When we encounter a cumulative sum value which we have already encountered before, which means that sub-array from the previous index+1 till the present index has equal number of 0’s and 1’s as they give a cumulative sum of 0 .

In a nutshell this problem is equivalent to finding two indexes i & j in array[] such that array[i] = array[j] and (j-i) is maximum. To store the first occurrence of each unique cumulative sum value we use a hash_map wherein if we get that value again we can find the sub-array size and compare it with the maximum size found till now.

Algorithm :

1. Let input array be arr[] of size n and max_size be the size of output sub-array.
2. Create a temporary array sumleft[] of size n. Store the sum of all elements from arr to arr[i] in sumleft[i].
3. There are two cases, the output sub-array may start from 0th index or may start from some other index. We will return the max of the values obtained by two cases.
4. To find the maximum length sub-array starting from 0th index, scan the sumleft[] and find the maximum i where sumleft[i] = 0.
5. Now, we need to find the subarray where subarray sum is 0 and start index is not 0. This problem is equivalent to finding two indexes i & j in sumleft[] such that sumleft[i] = sumleft[j] and j-i is maximum. To solve this, we create a hash table with size = max-min+1 where min is the minimum value in the sumleft[] and max is the maximum value in the sumleft[]. Hash the leftmost occurrences of all different values in sumleft[]. The size of hash is chosen as max-min+1 because there can be these many different possible values in sumleft[]. Initialize all values in hash as -1.
6. To fill and use hash[], traverse sumleft[] from 0 to n-1. If a value is not present in hash[], then store its index in hash. If the value is present, then calculate the difference of current index of sumleft[] and previously stored value in hash[]. If this difference is more than maxsize, then update the maxsize.
7. To handle corner cases (all 1s and all 0s), we initialize maxsize as -1. If the maxsize remains -1, then print there is no such subarray.

Pseudo Code:

```int sum_left[n]
Run a loop from i=0 to n-1
if(arr[i]==0)
sumleft[i] = sumleft[i-1]+-1
else
sumleft[i] = sumleft[i-1]+-1
if (sumleft[i] > max)
max = sumleft[i];

Run a loop from i=0 to n-1
if (sumleft[i] == 0)
{
maxsize = i+1;
startindex = 0;
}

// Case 2: fill hash table value. If already
then use it

if (hash[sumleft[i]-min] == -1)
hash[sumleft[i]-min] = i;
else
{
if ((i - hash[sumleft[i]-min]) > maxsize)
{
maxsize = i - hash[sumleft[i]-min];
startindex = hash[sumleft[i]-min] + 1;
}
}

return maxsize```

## C++

 `// C++ program to find largest subarray with equal number of``// 0's and 1's.` `#include ``using` `namespace` `std;` `// Returns largest subarray with equal number of 0s and 1s` `int` `maxLen(``int` `arr[], ``int` `n)``{``    ``// Creates an empty hashMap hM` `    ``unordered_map<``int``, ``int``> hM;` `    ``int` `sum = 0; ``// Initialize sum of elements``    ``int` `max_len = 0; ``// Initialize result``    ``int` `ending_index = -1;` `    ``for` `(``int` `i = 0; i < n; i++)``        ``arr[i] = (arr[i] == 0) ? -1 : 1;` `    ``// Traverse through the given array` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``// Add current element to sum` `        ``sum += arr[i];` `        ``// To handle sum=0 at last index` `        ``if` `(sum == 0) {``            ``max_len = i + 1;``            ``ending_index = i;``        ``}` `        ``// If this sum is seen before, then update max_len``        ``// if required` `        ``if` `(hM.find(sum) != hM.end()) {``            ``if` `(max_len < i - hM[sum]) {``                ``max_len = i - hM[sum];``                ``ending_index = i;``            ``}``        ``}``        ``else` `// Else put this sum in hash table``            ``hM[sum] = i;``    ``}` `    ``for` `(``int` `i = 0; i < n; i++)``        ``arr[i] = (arr[i] == -1) ? 0 : 1;` `    ``printf``(``"%d to %d\n"``,``           ``ending_index - max_len + 1, ending_index);` `    ``return` `max_len;``}` `// Driver method` `int` `main()``{``    ``int` `arr[] = { 1, 0, 0, 1, 0, 1, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``maxLen(arr, n);``    ``return` `0;``}` `// This code is contributed by Aditya Goel`

## C

 `// A O(n) program to find the largest subarray``// with equal number of 0s and 1s` `#include ``#include ` `// A utility function to get maximum of two``// integers` `int` `max(``int` `a, ``int` `b) { ``return` `a > b ? a : b; }` `// This function Prints the starting and ending``// indexes of the largest subarray with equal``// number of 0s and 1s. Also returns the size``// of such subarray.` `int` `findSubArray(``int` `arr[], ``int` `n)``{``    ``// variables to store result values` `    ``int` `maxsize = -1, startindex;` `    ``// Create an auxiliary array sunmleft[].``    ``// sumleft[i] will be sum of array``    ``// elements from arr to arr[i]` `    ``int` `sumleft[n];` `    ``// For min and max values in sumleft[]` `    ``int` `min, max;``    ``int` `i;` `    ``// Fill sumleft array and get min and max``    ``// values in it.  Consider 0 values in arr[]``    ``// as -1` `    ``sumleft = ((arr == 0) ? -1 : 1);``    ``min = arr;``    ``max = arr;``    ``for` `(i = 1; i < n; i++) {``        ``sumleft[i] = sumleft[i - 1]``                     ``+ ((arr[i] == 0) ? -1 : 1);``        ``if` `(sumleft[i] < min)``            ``min = sumleft[i];``        ``if` `(sumleft[i] > max)``            ``max = sumleft[i];``    ``}` `    ``// Now calculate the max value of j - i such``    ``// that sumleft[i] = sumleft[j]. The idea is``    ``// to create a hash table to store indexes of all``    ``// visited values.``    ``// If you see a value again, that it is a case of``    ``// sumleft[i] = sumleft[j]. Check if this j-i is``    ``// more than maxsize.``    ``// The optimum size of hash will be max-min+1 as``    ``// these many different values of sumleft[i] are``    ``// possible. Since we use optimum size, we need``    ``// to shift all values in sumleft[] by min before``    ``// using them as an index in hash[].` `    ``int` `hash[max - min + 1];` `    ``// Initialize hash table` `    ``for` `(i = 0; i < max - min + 1; i++)``        ``hash[i] = -1;` `    ``for` `(i = 0; i < n; i++) {``        ``// Case 1: when the subarray starts from``        ``// index 0` `        ``if` `(sumleft[i] == 0) {``            ``maxsize = i + 1;``            ``startindex = 0;``        ``}` `        ``// Case 2: fill hash table value. If already``        ``// filled, then use it` `        ``if` `(hash[sumleft[i] - min] == -1)``            ``hash[sumleft[i] - min] = i;``        ``else` `{``            ``if` `((i - hash[sumleft[i] - min]) > maxsize) {``                ``maxsize = i - hash[sumleft[i] - min];``                ``startindex = hash[sumleft[i] - min] + 1;``            ``}``        ``}``    ``}``    ``if` `(maxsize == -1)``        ``printf``(``"No such subarray"``);``    ``else``        ``printf``(``"%d to %d"``, startindex,``               ``startindex + maxsize - 1);` `    ``return` `maxsize;``}` `/* Driver program to test above functions */``int` `main()``{``    ``int` `arr[] = { 1, 0, 0, 1, 0, 1, 1 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSubArray(arr, size);``    ``return` `0;``}`

## Java

 `import` `java.util.HashMap;` `class` `LargestSubArray1 {` `    ``// Returns largest subarray with``    ``// equal number of 0s and 1s` `    ``int` `maxLen(``int` `arr[], ``int` `n)``    ``{``        ``// Creates an empty hashMap hM` `        ``HashMap hM``            ``= ``new` `HashMap();` `        ``// Initialize sum of elements``        ``int` `sum = ``0``;` `        ``// Initialize result``        ``int` `max_len = ``0``;``        ``int` `ending_index = -``1``;``        ``int` `start_index = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``arr[i] = (arr[i] == ``0``) ? -``1` `: ``1``;``        ``}` `        ``// Traverse through the given array` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``// Add current element to sum` `            ``sum += arr[i];` `            ``// To handle sum=0 at last index` `            ``if` `(sum == ``0``) {``                ``max_len = i + ``1``;``                ``ending_index = i;``            ``}` `            ``// If this sum is seen before,``            ``// then update max_len if required``            ``if` `(hM.containsKey(sum)) {``                ``if` `(max_len < i - hM.get(sum)) {``                    ``max_len = i - hM.get(sum);``                    ``ending_index = i;``                ``}``            ``}``            ``else` `// Else put this sum in hash table``                ``hM.put(sum, i);``        ``}` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``arr[i] = (arr[i] == -``1``) ? ``0` `: ``1``;``        ``}` `        ``int` `end = ending_index - max_len + ``1``;``        ``System.out.println(end + ``" to "` `+ ending_index);` `        ``return` `max_len;``    ``}` `    ``/* Driver program to test the above functions */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``LargestSubArray1 sub = ``new` `LargestSubArray1();``        ``int` `arr[] = { ``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1` `};``        ``int` `n = arr.length;` `        ``sub.maxLen(arr, n);``    ``}``}` `// This code has been by Mayank Jaiswal(mayank_24)`

## Python3

 `# Python 3 program to find largest``# subarray with equal number of``# 0's and 1's.` `# Returns largest subarray with``# equal number of 0s and 1s``def` `maxLen(arr, n):` `    ``# NOTE: Dictonary in python in``    ``# implemented as Hash Maps.``    ``# Create an empty hash map (dictionary)``    ``hash_map ``=` `{} ``    ``curr_sum ``=` `0``    ``max_len ``=` `0``    ``ending_index ``=` `-``1` `    ``for` `i ``in` `range` `(``0``, n):``        ``if``(arr[i] ``=``=` `0``):``            ``arr[i] ``=` `-``1``        ``else``:``            ``arr[i] ``=` `1` `    ``# Traverse through the given array``    ``for` `i ``in` `range` `(``0``, n):``    ` `        ``# Add current element to sum``        ``curr_sum ``=` `curr_sum ``+` `arr[i]` `        ``# To handle sum = 0 at last index``        ``if` `(curr_sum ``=``=` `0``):``            ``max_len ``=` `i ``+` `1``            ``ending_index ``=` `i` `        ``# If this sum is seen before,``        ``if` `curr_sum ``in` `hash_map:``            ` `            ``# If max_len is smaller than new subarray``            ``# Update max_len and ending_index``            ``if` `max_len < i ``-` `hash_map[curr_sum]:``                ``max_len ``=` `i ``-` `hash_map[curr_sum]``                ``ending_index ``=` `i``        ``else``:` `            ``# else put this sum in dictionary``            ``hash_map[curr_sum] ``=` `i ``        ` `    ``for` `i ``in` `range` `(``0``, n):``        ``if``(arr[i] ``=``=` `-``1``):``            ``arr[i] ``=` `0``        ``else``:``            ``arr[i] ``=` `1``            ` `    ``print` `(ending_index ``-` `max_len ``+` `1``, end ``=``" "``)``    ``print` `(``"to"``, end ``=` `" "``)``    ``print` `(ending_index)` `    ``return` `max_len` `# Driver Code``arr ``=` `[``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1``]``n ``=` `len``(arr) ` `maxLen(arr, n)``    ` `# This code is contributed``# by Tarun Garg`

## C#

 `// C# program to find the largest subarray``// with equal number of 0s and 1s``using` `System;``using` `System.Collections.Generic;` `class` `LargestSubArray1 {` `    ``// Returns largest subarray with``    ``// equal number of 0s and 1s``    ``public` `virtual` `int` `maxLen(``int``[] arr, ``int` `n)``    ``{` `        ``// Creates an empty Dictionary hM``        ``Dictionary<``int``,``                   ``int``>``            ``hM = ``new` `Dictionary<``int``,``                                ``int``>();` `        ``int` `sum = 0; ``// Initialize sum of elements``        ``int` `max_len = 0; ``// Initialize result``        ``int` `ending_index = -1;``        ``int` `start_index = 0;` `        ``for` `(``int` `i = 0; i < n; i++) {``            ``arr[i] = (arr[i] == 0) ? -1 : 1;``        ``}` `        ``// Traverse through the given array``        ``for` `(``int` `i = 0; i < n; i++) {``            ``// Add current element to sum``            ``sum += arr[i];` `            ``// To handle sum=0 at last index``            ``if` `(sum == 0) {``                ``max_len = i + 1;``                ``ending_index = i;``            ``}` `            ``// If this sum is seen before,``            ``// then update max_len``            ``// if required``            ``if` `(hM.ContainsKey(sum)) {``                ``if` `(max_len < i - hM[sum]) {``                    ``max_len = i - hM[sum];``                    ``ending_index = i;``                ``}``            ``}` `            ``else` `// Else put this sum in hash table``            ``{``                ``hM[sum] = i;``            ``}``        ``}` `        ``for` `(``int` `i = 0; i < n; i++) {``            ``arr[i] = (arr[i] == -1) ? 0 : 1;``        ``}` `        ``int` `end = ending_index - max_len + 1;``        ``Console.WriteLine(end + ``" to "` `+ ending_index);` `        ``return` `max_len;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{` `        ``LargestSubArray1 sub = ``new` `LargestSubArray1();``        ``int``[] arr = ``new` `int``[] {``            ``1,``            ``0,``            ``0,``            ``1,``            ``0,``            ``1,``            ``1``        ``};` `        ``int` `n = arr.Length;``        ``sub.maxLen(arr, n);``    ``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output:

`0 to 5`

Complexity Analysis:

• Time Complexity: O(n).
As the given array is traversed only once.
• Auxiliary Space: O(n).
As hash_map has been used which takes extra space.